An electronics store sells 60 entertainment systems per month at $800 each. For every $20 increase in the selling price, the store sells one fewer system. What should the store set the price to in order to maximize the revenue? What is the maximum revenue? Provide an algebraic solution for full marks.
Let the number of price increases by x. Then since revenue = price * quantity,
r(x) = (800+20x)(60-x)
The vertex of that parabola is at (10,50000)
So max revenue of $50K when the price is 800+20*10 = $1000
Well, if we analyze the problem, we see that as the selling price increases, the number of systems sold decreases. This is like a seesaw but with money and entertainment systems. Now, let's give it a whirl and solve the problem!
Let "x" be the number of $20 price increases.
So, if the initial selling price is $800, then the new selling price will be $800 + $20x.
The number of systems sold will be 60 - x because for every $20 increase, one less system is sold.
Now, let's calculate the revenue:
Revenue = Selling Price * Number of Systems Sold
Revenue = ($800 + $20x) * (60 - x)
To maximize the revenue, we have to find the value of "x" that gives us the maximum revenue. To do that, we need to find the vertex of the quadratic equation that represents the revenue.
The quadratic equation for revenue is:
R(x) = ($800 + $20x) * (60 - x)
Expanding it:
R(x) = 48000 - 800x + 1200x - 20x^2
Simplifying:
R(x) = -20x^2 + 400x + 48000
Now, we can use the formula for the x-coordinate of the vertex:
x = -b / 2a
Where a = -20 and b = 400:
x = -400 / (2 * -20)
x = -400 / -40
x = 10
So, the number of $20 price increases should be 10 to maximize the revenue.
Now, let's substitute x = 10 back into the equation for the selling price to find the maximum revenue:
Selling Price = $800 + $20x
Selling Price = $800 + $20(10)
Selling Price = $800 + $200
Selling Price = $1000
Therefore, the store should set the price to $1000 to maximize the revenue.
To find the maximum revenue, let's substitute x = 10 back into the equation for revenue:
R(x) = -20x^2 + 400x + 48000
R(10) = -20(10)^2 + 400(10) + 48000
R(10) = -20(100) + 4000 + 48000
R(10) = -2000 + 4000 + 48000
R(10) = 50000
Therefore, the maximum revenue is $50,000.
So, the store should set the price to $1000 to maximize the revenue, and the maximum revenue would be $50,000. I hope that adds some "priceless" humor to the solution!
To find the price that will maximize revenue, we need to determine the relationship between the selling price and the number of systems sold.
Let's assume that the selling price is denoted by x and the number of systems sold is denoted by y.
Given that the store currently sells 60 entertainment systems per month at $800 each and for every $20 increase in the selling price, the store sells one fewer system, we can create an equation for the relationship between the price and the quantity sold:
x = 800 + (60 - y) * 20
The revenue is given by the product of the selling price and the quantity sold:
R = x * y
Substituting the equation for x obtained above:
R = (800 + (60 - y) * 20) * y
To find the price that maximizes revenue, we need to find the value of y that maximizes the equation for R. We can do this by differentiating R with respect to y, setting the derivative equal to zero, and solve for y:
dR/dy = (800 + (60 - y) * 20) + (800 - 20y)
dR/dy = 1600 - 40y
Setting dR/dy equal to zero:
1600 - 40y = 0
40y = 1600
y = 40
Substituting y = 40 back into the equation for x:
x = 800 + (60 - 40) * 20
x = 800 + 20 * 20
x = 800 + 400
x = 1200
Therefore, the store should set the price to $1200 in order to maximize revenue.
To find the maximum revenue, substitute the values of x and y into the equation for R:
R = (800 + (60 - 40) * 20) * 40
R = 1200 * 40
R = 48000
Therefore, the maximum revenue is $48,000.
To maximize the revenue, we need to find the selling price at which the product of the number of units sold and the selling price is maximum. Let's go step by step to find the solution.
Step 1: Define the variables:
Let x be the number of $20 price increases (increments) from the initial selling price.
So, the selling price can be expressed as 800 + (x * 20).
Step 2: Determine the number of units sold:
For every $20 increase in selling price, one fewer system is sold.
Therefore, the number of units sold can be expressed as 60 - x.
Step 3: Calculate the revenue:
The revenue can be calculated by multiplying the number of units sold by the selling price:
Revenue = (60 - x) * (800 + (x * 20))
Simplifying this expression, we have:
Revenue = (60 - x) * (800 + 20x)
= 48000 + 1200x - 800x - 20x^2
= -20x^2 + 400x + 48000
Step 4: Find the maximum revenue:
To find the maximum revenue, we need to find the value of x that maximizes the revenue function.
The revenue function is quadratic in form, and the maximum or minimum value occurs at the vertex of the parabola. The x-coordinate of the vertex can be calculated using the formula: x = -b / 2a.
In our equation, a is -20, and b is 400. Substituting these values into the formula, we have:
x = -400 / (2 * -20)
x = -400 / -40
x = 10
Therefore, the maximum revenue occurs when x = 10.
Step 5: Calculate the selling price and maximum revenue:
Substituting x = 10 into our selling price expression:
Selling price = 800 + (10 * 20)
= 800 + 200
= 1000
Substituting x = 10 into our revenue expression:
Revenue = -20(10)^2 + 400(10) + 48000
= -20(100) + 4000 + 48000
= -2000 + 4000 + 48000
= 50000
Therefore, the store should set the selling price to $1000 in order to maximize the revenue, and the maximum revenue will be $50,000.